HDU 2141 Can you find it? (二分查找)

Problem Description

Give you three sequences of numbers A, B, C, then we give you a number X. Now you need to calculate if you can find the three numbers Ai, Bj, Ck, which satisfy the formula Ai+Bj+Ck = X.



Input

There are many cases. Every data case is described as followed: In the first line there are three integers L, N, M, in the second line there are L integers represent the sequence A, in the third line there are N integers represent the sequences B, in the forth
line there are M integers represent the sequence C. In the fifth line there is an integer S represents there are S integers X to be calculated. 1<=L, N, M<=500, 1<=S<=1000. all the integers are 32-integers.



Output

For each case, firstly you have to print the case number as the form "Case d:", then for the S queries, you calculate if the formula can be satisfied or not. If satisfied, you print "YES", otherwise print "NO".



Sample Input

3 3 3
1 2 3
1 2 3
1 2 3
3
1
4
10

Sample Output

Case 1:
NO
YES
NO

#include <iostream>

#include <string.h>

#include <algorithm>

#include <stdio.h>

using namespace std;

bool Search(long long Sum[],int low,int high,int value)

{

while(low<=high)

{

int mid=(low+high)/2;

if(Sum[mid]==value)return true;

else if(Sum[mid]<value)low=mid+1;

else high=mid-1;

}

return false;

}

int main()

{

int cases=0;

int L,N,M;

while(cin>>L>>N>>M)

{

int A[510]={0},B[510]={0},C[510]={0},S,X;

long long Sum[250010]={0},len=0;

for(int i=0;i<L;i++)

cin>>A[i];

for(int i=0;i<N;i++)

cin>>B[i];

for(int i=0;i<M;i++)

cin>>C[i];

for(int i=0;i<N;i++)//计算A与B所有元素之和

for(int j=0;j<M;j++)

Sum[len++]=A[i]+B[j];

sort(Sum,Sum+len);//排序以便二分查找

sort(C,C+M);

cout<<"Case "<<++cases<<":"<<endl;

cin>>S;

while(S--)

{

cin>>X;

if(Sum[0]+C[0]>X||Sum[len-1]+C[M-1]<X)

{

cout<<"NO"<<endl;

continue;

}

int flag=0;

for(int i=0;i<M;i++)

{

int temp=X-C[i];

if(Search(Sum,0,len-1,temp)==true)//一旦找到就OK

{

cout<<"YES"<<endl;

flag=1;

break;

}

}

if(flag==0)

cout<<"NO"<<endl;

}

}

return 0;

}